Arc length and chord only.

Question

Hello everyone hope you are all well.

Is it possible to find the radius of a circle if the only information you have is arc length and its associated chord?

Thank you all for your time.

Answer 1

Starting from @orangeskid's answer, you can almost have the analytical solution of the equation $$\frac{c}{a} = \frac{\sin(\frac{\theta}{2})}{\frac{\theta}{2}}$$ Let $k=\frac{c}{a}$ and $x=\frac{\theta}{2}$ to make $$k x=\sin(x)$$ Now, for a first approximation, use the $1,400$ years old approximation $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad \qquad(0\leq x\leq\pi)$$ which gives $$x_0=\frac{\pi k-4+2 \sqrt{-\pi ^2 k^2+2 \pi k+4}}{2 k}$$ and now, perform one single iteration of Newton method $$x_1=\frac{\sin (x_0)-x_0 \cos (x_0)}{k-\cos (x_0)}$$ Some results $$\left( \begin{array}{cccc} k & x_0 & x_1 & \text{solution} \\ 0.05 & 2.99294 & 2.99146 & 2.99146 \\ 0.10 & 2.85371 & 2.85234 & 2.85234 \\ 0.15 & 2.72180 & 2.72114 & 2.72114 \\ 0.20 & 2.59560 & 2.59574 & 2.59574 \\ 0.25 & 2.47378 & 2.47458 & 2.47458 \\ 0.30 & 2.35522 & 2.35644 & 2.35644 \\ 0.35 & 2.23894 & 2.24033 & 2.24033 \\ 0.40 & 2.12403 & 2.12535 & 2.12535 \\ 0.45 & 2.00960 & 2.01068 & 2.01067 \\ 0.50 & 1.89477 & 1.89549 & 1.89549 \\ 0.55 & 1.77859 & 1.77894 & 1.77894 \\ 0.60 & 1.65996 & 1.66003 & 1.66003 \\ 0.65 & 1.53760 & 1.53761 & 1.53761 \\ 0.70 & 1.40988 & 1.41019 & 1.41019 \\ 0.75 & 1.27459 & 1.27570 & 1.27570 \\ 0.80 & 1.12856 & 1.13111 & 1.13110 \\ 0.85 & 0.96677 & 0.97138 & 0.97135 \\ 0.90 & 0.78035 & 0.78676 & 0.78668 \\ 0.95 & 0.55016 & 0.55192 & 0.55191 \end{array} \right)$$

Answer 2

The problem is interesting. Notice that if the angle at center is $\theta$ ($0\le \theta \le 2\pi$) then $$c = 2 R \sin \frac{\theta}{2}\\ a = R \theta$$

From here we get $$\frac{c}{a} = \frac{\sin\frac{\theta}{2}}{\frac{\theta}{2}}$$

Now, on the interval $[0, \pi]$ the function $\phi \to \frac{\sin\phi}{\phi}$ is strictly decreasing, with image $[0, 1]$.

Therefore, knowing $c$, $a$, get $\frac{c}{a}$, then $\phi\colon = \frac{\theta}{2}$, and then $R$.

Example: $c=10$, $a=12$. With $\phi\colon = \frac{\theta}{2}$ we get $$\frac{\sin \phi}{\phi} = \frac{10}{12}$$ and so $\phi= 1.0267\ldots$ and $$R = \frac{a}{2 \phi} = \frac{12}{2 \cdot 1.0267\ldots} = 5.8437\ldots$$

Note: this allows calculating the radius of a circle when we have access only to a portion of it.

Answer 3

What have you tried?

HINT

Can you draw a figure of chord and arc?

$$ s= R \theta;\quad L= 2 R \sin \theta/2; $$

A series approximation is computable to any degree of approximation depending on the number of terms chosen in sine series

$$ \dfrac{L}{2R}= \dfrac{s}{2R}-\dfrac{(s/2R)^3}{3!} +\dfrac{(s/2R)^5}{5!} - ...$$